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3 years ago

5

Is 2/5 equivalent to 0.25

2 answers:

spayn [35]3 years ago

4 0

I would say no. 2/5 =. 40
. 25 = ¼
Imagine you have 5 parts.
1 = 20% each
I hope this helps.

notsponge [240]3 years ago

3 0

No it isn't because 2/5=0.4

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Standardized tests: In a particular year, the mean score on the ACT test was and the standard deviation was . The mean score on

Scorpion4ik [409]

Complete question :

Standardized tests: In a particular year, the mean score on the ACT test was 19.3 and the standard deviation was 5.3. The mean score on the SAT mathematics test was 532 and the standard deviation was 128. The distributions of both scores were approximately bell-shaped. Round the answers to at least two decimal places. Part: 0/4 Part 1 of 4 (a) Find the z-score for an ACT score of 26. The Z-score for an ACT score of 26 is

Answer:

1.26

Step-by-step explanation:

Given that:

For ACT:

Mean score, m = 19.3

Standard deviation, s = 5.3

Zscore for ACT score of 26;

Using the Zscore formula :

(x - mean) / standard deviation

x = 26

Zscore :

(26 - 19.3) / 5.3

= 6.7 / 5.3

= 1.2641509

= 1.26

6 0

3 years ago

The sum of the cube of a number and 12

Black_prince [1.1K]

Let the "number" be n.

The expression would be $n^3+12$

3 0

3 years ago

Help asap please!<br><br> 2/5+????=7/10<br><br> Find the missing fraction

docker41 [41]

Answer:

3/10

Step-by-step explanation:

2/5 + x = 7/10

x = 7/10 - 2/5

x = 7/10 - 4/10

x = 3/10

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2 years ago

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Let A and B be events with =PA0.7, =PB0.3, and =PA or B0.9. (a) Compute PA and B. (b) Are A and B mutually exclusive? Explain. (

kvasek [131]

Answer:

a) $P(A \cup B) = P(A) +P(B) - P(A\cap B)$

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

b) False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

c) False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

Step-by-step explanation:

For this case we have the following probabilities given:

$P(A) = 0.7, P(B) =0.7, P(A \cup B) =0.9$

Part a

We want to calculate the following probability: $P(A \cap B)$

And we can use the total probability rule given by:

$P(A \cup B) = P(A) +P(B) - P(A\cap B)$

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

Part b

False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

Part c

False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

4 0

4 years ago

Which statement is sometimes true will give brainlist if righ and 10 extra points

arlik [135]

A rectangle is a square.

4 0

3 years ago

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